UNIVERSITY OF BRISTOL. Examination for the Degrees of B.Sc. and M.Sci. (Level C/4) ANALYSIS 1B, SOLUTIONS MATH (Paper Code MATH-10006)

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1 UNIVERSITY OF BRISTOL Exmitio for the Degrees of B.Sc. d M.Sci. (Level C/4) ANALYSIS B, SOLUTIONS MATH 6 (Pper Code MATH-6) My/Jue 25, hours 3 miutes This pper cotis two sectios, A d B. Plese use seprte swer book for ech sectio. Sectio A cotis 5 short questios, ALL of which you should swer. Sectio B hs 2 loger questios, of which you should swer ALL. Sectio A cotributes 4% of the mrks d Sectio B cotributes 6% of the mrks for this pper. Clcultors re ot permitted i this exmitio. O this exmitio, the mrkig scheme is idictive d is iteded oly s guide to the reltive weightig of the questios. Pge of 8 Do ot tur over util istructed.

2 Cot... Alysis B-5 SOLUTIONS Sectio A: Short Questios A. (8 mrks) Cosider fuctio f : (, b) R. () Defie wht it mes for f to be differetible t x (, b) d give the defiitio of the derivtive. (b) Show tht if f is differetible t x (, b), the f is cotiuous t x. Solutio (bookwork): f(y) f(x) () f is differetible t x if the limit lim y x y x the derivtive of f t x, f f(y) f(x) (x) = lim y x. y x exists, d the the limit is clled (b) Usig f(y) f(x) = f(y) f(x) (y x), for y x, we get lim y x y x (f(y) f(x) = ) limy x (y x) = f (x) =, sice f is differetible t x. ( limy x f(y) f(x) y x A2. (8 mrks) () Stte the Me Vlue Theorem. (b) Show tht if f : [, b] R is cotiuous d differetible o (, b) d if there exists costt M > such tht f (x) M for ll x (, b), the for ll x, y [, b] f(y) f(x) M y x. (c) Show tht if f (x) = for ll x (, b), the f is costt. Solutio ((), (c) bookwork, (b) similr to homework): () Let f : [, b] R be cotiuous d differetible o (, b), the there exists c (, b) such tht f(b) f() = (b )f (c) (b) If f (x) M for ll x (, b), by the me vlue theorem pplied to f : [x, y] R there exists c (x, y) such tht f(y) f(x) = (y x)f (c) d hece f(y) f(x) M y x. (c) Let δ >. Sice f (x) δ for ll x [, b], prt (b) gives tht for ll y, x [, b], f(y) f(x) δ y x δ(b ). Sice this holds for ll δ > we c coclude tht f(y) f(x) d so f(y) = f(x). Hece f is costt o [, b]. A3. (8 mrks) Pge 2 of 8

3 () Let ( ) N be bouded sequece of rel umbers d set c := sup{ m ; m }. Show tht the sequece (c ) N coverges d express lim sup, i terms of c. (b) Compute c for the sequece = ( ) ( ), d thus fid lim sup. Solutio (() bookwork, (b) similr to see exmples ): () Sice for k we hve { m ; m k} { m ; m } it follows tht sup{ m ; m k} sup{ m ; m }, i.e, c k c, so the sequece (c ) is mootoe decresig. Ad sice ( ) is bouded, (c ) is bouded, too, hece by the mootoe covergece theorem (c ) is coverget. The limit superior of ( ) is defied s the limit of c, lim sup = lim c. (b) Sice / 2 is decresig we hve for eve c = ( ) 3+ d for odd 2 c = ( ) 3+ (+) 2 d sice lim = lim 2 = we fid lim (+ 2 c = 3. So lim sup = 3. A4. (8 mrks) Let A R d f, f : A R () Wht does it me tht sequece of fuctios (f ) N coverges poitwise to fuctio f? (b) Wht does it me tht sequece of fuctios (f ) N coverges uiformly to fuctio f? (c) Stte (without proof) Weierstrss s theorem o uiform covergece. Solutio (bookwork): () For y x A, the sequece (f (x)) N coverges to f(x), or, for y x A d ε > there exists N N such tht for ll N f(x) f (x) ε. (b) For y ε > there exists N N such tht for ll N d ll x A f(x) f (x) ε. (c) Let A R d let (f ) N be sequece o cotiuous fuctios which coverge uiformly to fuctio f : A R. The f is cotiuous o A. A5. (8 mrks) () Defie wht it mes to sy tht the power series = x hs rdius of covergece R i ech of the cses (i) R = (ii) < R < (iii) R = Pge 3 of 8

4 (b) Determie the rdius of covergece of Solutio: = 5 (!) 2 (2)! x. () (i) R = mes the series is diverget for y x. (ii) < R < mes the series is coverget for x ( R, R) d diverget for x (, R) (R, ). (iii) R = mes the series is coverget for y x R. (b) We use the rtio test: + = 5+ (( + )!) 2 (2)! (2 + 2)! 5 (!) = 5( + ) 2 2 (2 + 2)(2 + ), so lim + = 5/4 d R = 4/5. Pge 4 of 8 Cotiued...

5 Cot... Alysis B-5 SOLUTIONS Sectio B: Loger Questios B. I this questio you c use without proof tht s(x) := si(x) d c(x) := cos(x) re differetible fuctios with s = c, c = s, s() =, c() = d sup x R s(x) = sup x R c(x) =. () (4 mrks) Write dow the first 4 terms i the Tylor series of f(x) = A cos(x) + B si(x) roud x =. Here A, B R re costt. (b) (8 mrks) Prove, by iductio o, tht if f : R R hs cotiuous th derivtive, the for x > where R (x) = f(x) = ( )! f (k) () k! x k + R (x) (x t) f () (t) dt (c) (4 mrks) Show tht if f : R R hs cotiuous th derivtive, the for x >, R (x) x! (d) (4 mrks) Show tht for y N d x > cos(x) ( ) k (2k)! x2k x 2 (2)! (e) ( mrks) Show tht for x > sup f () (t). t [ x,x] si(x) ( ) k (2k + )! x2k+ cos(x) 2 x x4 d si(x) x 6 x3, x 2+ (2 + )!. d derive from these estimtes tht cos(x) hs exctly oe zero i the itervl [, 2]. Stte crefully which results you use. Solutio ((() similr to homework, (b) (c) bookwork, (d) homework, (e) doe i clss, but chllegig): () (b) The cse = is A + Bx A 2 x2 B 6 x3 + A 24 x4 f(x) = f() + f (t) dt which is the Fudmetl Theorem of Clculus. So suppose the results holds for, d ssume f hs ctully cotiuous + st derivtive, the f(x) = f (k) () k! Pge 5 of 8 x k + R (x)

6 d itegrtio by prts gives hece R (x) = ( )! (x t) f () (t) dt =! (x t) f () (t) x +! = f () ()! f(x) = x + R + (x), f (k) () k! x k + R + (x). (x t) f (+) (t) dt (c) We use tht for regulr fuctio f we hve f dx f dx to get R (x) (x t) f () (t) dt ( )! sup f () x (t) (x t) dt t [,x] ( )! = sup f () (t) x t [,x]!. Altertively o c use the Me Vlue Theorem for itegrls. (d) We use prt (b) d (c) with the stdrd properties of the sie d cosie. For f(x) = si x we hve f (2) (x) = ( ) si(x), hece f (2) () = d f (2+) (x) = ( ) cos(x), hece f (2+) () = ( ). Ad with f () (t) we obti si(x) = ( ) k (2k + )! x2k+ + R 2+ (x) with R 2+ (x) x2+ (2+)!. Similrly with g(x) = cos(x) we hve g (2) (x) = ( ) cos(x) d g (2+) (x) = ( ) si(x), hece g (2) () = ( ), d g (2+) () = d g () (t), so cos(x) = ( ) k (2k)! x2k + R 2 (x) with R 2 (x) x2 (2)!. (e) Usig prt (d) with = for the sie d = 2 for the cosie gives si(x) x 6 x3, cos(x) 2 x2 + 4! x4. Now we hve cos() = d cos(2) = + 2 = <. Sice cos(x) is 4! 3 3 cotiuous, by the itermedite vlue theorem there must be t lest oe x (, 2) such tht cos(x) =. Now the derivtive of cos(x) is si(x) d we hve si(x) x + 6 x3 = x ( 6 ) x2 d sice for x [, 2] we hve 6 x2 2 2 /6 = 2/3 = /3 it follows tht cos(x) x/3 for x [, 2]. So cos(x) is strictly decresig o [, 2] d c therefore hve t most oe zero. Pge 6 of 8

7 B2. () (i) (4 mrks) Defie wht it mes for ψ : [, b] R to be step fuctio, d defie the itegrl ψ(x) dx. (You do ot hve to show tht the itegrl is idepedet of the prtitio.) (ii) (4 mrks) Show tht ψ(x)dx (b ) ψ. (iii) (4 mrks) Suppose ψ d φ re two sequeces of step fuctios with lim ψ φ =, show tht lim ψ (x) dx φ (x) dx =,. (You c use without proof lierity of the itegrl d tht ψ φ re step fuctios.) (iv) (8 mrks) Suppose f : [, b] R d there exists sequece of step fuctios ψ : [, b] R, N, with lim ψ f =. Defie the itegrl of f, i.e. show tht the sequece ψ (x) dx coverges, d tht the limit is idepedet of the choice of the pproximtig sequece. (b) ( mrks) Let f : [, ) be positive mootoic decresig fuctio. Solutio: (i) Show tht for y N (ii) Show tht f( + ) N f() =2 + N f(x) dx f(). f(x) dx N = d tht = f() coverges if d oly if lim y f(), y f(x) dx exists. () (i) ψ : [, b] R is step fuctio if there exists prtitio P = {x =, x,, x = b} of [, b] such tht ψ is costt o ech itervl (x k, x k ), k =, 2,,. Let the vlue of ψ o (x k, x k ) be c k, the the itegrl of ψ is defied s ψ(x) dx := c k (x k x k ). (ii) we hve c k ψ for k =, 2,,, d hece c k (x k x k ) c k (x k x k ) ψ (b ) k= where we used k= (x k x k ) = b. (iii) We hve ψ dx φ dx = ψ φ dx, hece with prt (i) ψ dx φ dx (b ) ψ φ. So if lim ψ φ =, the lim ψ dx φ dx =. k= Pge 7 of 8 k=

8 (iv) We defie f(x) dx := lim ψ (x) dx. Notice tht sice ψ is Cuchy sequece d ψ dx ψ m dx (b ) ψ ψ m, we hve tht ψ dx is Cuchy sequece, too. So the sequece coverges. Now if φ is other sequece of step fuctios covergig uiformly to f, the ψ φ coverges uiformly to, d sice ψ dx φ dx (b ) ψ φ, we hve lim ψ (x) dx = lim φ (x) dx. (b) (i) Sice f is decresig we hve for x [, + ] tht f() f(x) f( + ), hece f() = + f() dx + (ii) We use tht N f(x) dx = N = result of (i) N f() =2 f(x) dx + N + f( + ) dx = f( + ). f(x) dx which gives us together with the f(x) dx N = y f(). So if the sum N = f() coverges, the lim y f(x) dx coverges, d if the sum diverges to + the y f(x) dx diverges to + s y. Sice f is positive these re the oly two possibilities. Pge 8 of 8 Ed of exmitio.